Question

The escape speed of a projectile on the earth’s surface is 11.2 km s–1. A body is projected out with thrice this speed. What is the speed of the body far away from the earth? Ignore the presence of the sun and other planets.​

Answer 1

Explanation:

Escape velocity of a projectile from the Earth, vesc = 11.2 km/s

Projection velocity of the projectile, vp = 3vesc

Mass of the projectile = m

Velocity of the projectile far away from the Earth = vf

Total energy of the projectile on the Earth = (1/2)mvp2 - (1/2)mvesc2

Gravitational potential energy of the projectile far away from the Earth is zero.

Total energy of the projectile far away from the Earth = (1/2)mvf2

From the law of conservation of energy, we have

(1/2)mvp2 - (1/2)mvesc2 = (1/2)mvf2

vf = ( vp2 - vesc2 )1/2

= [ (3vesc)2 - vesc2 ]1/2

= √8 vesc

= √8 × 11.2 = 31.68 km/s.

Answer 2

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Given ,

escape speed on the earth's surface (Ve) = 11.2 km/s

speed of projection of the body (v) = 3Ve = 3×11.2 = 33.6 km/s

Let v and v' be the speed of the body at the time of projection and at a point far from the earth.

Initial K.E of the body = 1/2mv²

Initial Gravitational PE of the body = -GMem/Re

Where, Me , Re are the mass , radius respectively.

At very far from the earth surface .

KE of the body = 1/2mv²

Gravitational PE of the body = 0

A/c to law of conservation of energy .

Total energy at the point of projection = total energy at very far from the earth's surface.

1/2mv² + (-GMem/Re) = 1/2mv'²

1/2 mv'² = 1/2mv² - GMem/Re

If Ve is the escape velocity , then,

1/2 mVe² = GMem/Re

1/2mv'² = 1/2mv² - 1/2mVe²

v'² = v - Ve²

= (3Ve)² - Ve²

= 8Ve²

v'² = 8Ve²

v' = 2√2Ve

v = 2 × 1.414 × 11.2 km/s

= 31.68 km/s

Hence, speed of the body at far away from the earth = 31.68 km/s